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Triangular Form of a matrix

  1. Nov 28, 2008 #1
    I am confused on how to find a matrix B in triangular form for some linear transformation T over a basis [tex] \{v_1,v_2, v_3\} [/tex].

    Suppose we are given a minimal polynomial [tex]m(x) = (x+1)^2 (x-2)[/tex].

    Do I want to find a basis [tex] \{w_1,w_2\} [/tex] for [tex]null(T+1)^2[/tex] such that [tex](T+1) w_1 = 0[/tex] and [tex](T+1) w_2 \in S(w_1)[/tex]? Is this because [tex](x+1)^2[/tex] has degree two? This is the part I'm not sure about.

    For [tex]w_3[/tex], should I just let it be a basis for [tex]null(T-2)[/tex]?

    I tried this for a specific transformation T and got the correct matrix B. (I checked the work by computing the matrix S that relates the old basis (v's) to the new basis (w's) and used the relation [tex]B = S^{-1} A S[/tex] where A is the matrix of T.)

    Thanks for the help!
     
  2. jcsd
  3. Nov 29, 2008 #2

    morphism

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    What is S?

    The general idea is as follows. If you take a basis for null(T+1)^2 and a basis for null(T-2), then their union will be a basis for your space V with respect to which T is block diagonal. In this case the first block will be 2x2 and the second one will be 1x1 (with just a '2' in it). So to make sure that T becomes upper triangular, we're going to have to see to it that the 2x2 block is upper triangular. One way to do this is to let w_1 be an eigenvector for T corresponding to -1 (i.e. pick a w_1 in null(T+1)) and then extend {w_1} to a basis {w_1, w_2} for null(T+1)^2. This choice of w_1 ensures that we get the 0s that we want on the first column.
     
    Last edited: Nov 29, 2008
  4. Nov 29, 2008 #3
    In [tex](T+1) w_2 \in S(w_1)[/tex] S means subspace. (I know it is confusing when S is also used for the matrix relating the two basis.)

    The space V is a direct sum of V1 = null(T+1)^2 and V2 = null(T-2). Is this the reason for the 2x2 matrix block, the diagonals are all -1 and the 1x1 matrix block, the diagonal is 2?

    To make the example concrete, suppose the matrix corresponding to the transformation T is
    [tex]A = \begin{bmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 0 & 0 & -1 \end{bmatrix}[/tex].

    A vector in null(A + I) is x = v1 - v2 (just by solving the equation (A + I)x=0 for x). Likewise, a vector in null(A + I)^2 is y = v2-v3.

    This gives me a new basis w1 = v1 - v2, w2 = v2 - v3, and w3 = v3. I checked this and it works. However, another vector in null(A + I)^2 is y'= v3-v2, so I'd get a different set of w1, w2, and w3 and I didn't get the matrix in block form as before. Shouldn't I get still get a matrix in diagonal form (but with different number in the upper triangular block)?

    I don't see how choosing a vector w1 such that T(w1) = -w1 will also guarantee the matrix for the eigenvalue -2 will also be in block form.
     
  5. Nov 30, 2008 #4

    morphism

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    No. This is the reason that we can write T as a direct sum of a 2x2 block and a 1x1 block with respect to this decomposition of V. Of course the 1x1 block will have to be a '2', because Tx=2x for vector in V2.

    I don't really follow what you're doing in the rest of the post. If the minimal polynomial of A is m(x)=(x+1)^2(x-1) (I haven't checked), then A won't be diagonalizable. But it will be upper triangular if you choose a good basis for null(A+1)^2. Think about why picking an eigenvector is a good idea.
     
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